Formula for Nim

Date: 02/22/2002 at 15:20:56
From: Kentaro Miura
Subject: NIM
Hi.
I have been attempting to figure out a formula for the game called
the Chip Game. You can play with any number of chips. Each player
takes turns to eliminate up to three coins at a time. The person to
take the last chip loses the game. (For example, there are 5 chips.
The first person takes 1 chip. His opponent takes 3 chips. The person
that took the first one lost.)
I have gained the ability to win some games. However, I would like to
know a complete formula that applies for any number.
Thank you.

Date: 02/22/2002 at 23:23:03
From: Doctor Twe
Subject: Re: NIM
Hi Kentaro - thanks for writing to Dr. Math.
This is one version of a popular mathematical game commonly called
NIM. If you type "nim" (without the quotation marks) and select
"complete words only" in our Ask Dr. Math archive searcher at:
http://mathforum.org/mathgrepform.html
You'll find a half dozen answers relating to the game of NIM and its
variants.
Let's start with a simple game. Assume we play first, and that our
opponent will play his best possible move each time.
Suppose we start with 4, 3, or 2 coins. What is our strategy? Simple;
take all but 1 coin. Our opponent has to take the last coin and lose.
Now suppose that we start with 5 coins.
* If we take 1 coin (leaving 4), our opponent will take 3 - we lose.
* If we take 2 coins (leaving 3), our opponent will take 2 - we lose.
* If we take 3 coins (leaving 2), our opponent will take 1 - we lose.
No matter what we do at 5 coins, we lose.
Next, suppose we start with 6 coins.
* If we take 1 coin, we leave our opponent with 5 - which we showed
was a losing position for him.
* If we take 2 coins (leaving 4), our opponent will take 3 - we lose.
* If we take 3 coins (leaving 3), our opponent will take 2 - we lose.
So our winning strategy is to take 1 coin, leaving our opponent with
the losing position of 5 coins.
Suppose we start with 7 coins.
* If we take 1 coin (leaving 6), our opponent will take 1 - leaving
us with 5 - we lose.
* If we take 2 coins, we leave our opponent with 5 - a losing
position.
* If we take 3 coins (leaving 4), our opponent will take 3 - we lose.
So our winning strategy is to take 2 coins, leaving our opponent with
the losing position of 5 coins.
Likewise if we start with 8 coins.
* If we take 1 coin (leaving 7), our opponent will take 2 - leaving
us with 5 - we lose.
* If we take 2 coins (leaving 6), our opponent will take 1 - leaving
us with 5 - we lose.
* If we take 3 coins, we leave our opponent with 5 - a losing
position.
So our winning strategy is to take 3 coins, leaving our opponent with
the losing position of 5 coins.
Now, suppose we start with 9 coins.
* If we take 1 coin (leaving 8), our opponent will take 3 - we lose.
* If we take 2 coins (leaving 7), our opponent will take 2 - we lose.
* If we take 3 coins (leaving 6), our opponent will take 1 - we lose.
No matter what we do at 9 coins, we lose.
Suppose we start with 10 coins.
* If we take 1 coin, we leave our opponent with 9 - which we showed
was a losing position.
* If we take 2 coins (leaving 8), our opponent will take 3 - we lose.
* If we take 3 coins (leaving 7), our opponent will take 2 - we lose.
So our winning strategy is to take 1 coin, leaving our opponent with
the losing position of 9 coins.
We can continue this, but let's put it in chart form:
# Coins We take Leaving
------- ------- -------
1 1 0 We lose.
2 1 1 Opponent must take it and lose.
3 2 1
4 3 1
5 X X Whatever we take, we lose.
6 1 5 Opponent will lose.
7 2 5
8 3 5
9 X X Whatever we take, we lose.
10 1 9 Opponent will lose.
11 2 9
: : :
Do you see the pattern? Every fourth number -- 1, 5, 9, 13, ... - is a
losing number. We want to leave our *opponent* with that many coins at
the start of his turn. To do that, we must get to the nearest losing
number, and then match his every move. If he takes N coins, we take
4-N to get to the next losing number.
For example, suppose we start with 20 coins. The nearest losing number
is 17, so we take 3. Suppose our opponent takes 2, leaving 15. We then
take 4-2 = 2 to get to the next losing number, which is 13. Suppose he
next takes 3, leaving 10. We then take 4-3 = 1 to get to the next
losing number, which is 9. Suppose he next takes 1, leaving 8. We then
take 4-1 = 3 to get to the next losing number, which is 5. At this
point, he realizes that he can't win, but takes 1 as a formality,
leaving us 4. We take 3, and he must take the last one - we win!
Try a few games, and then see if you can find the winning strategy for
the following variants:
* Each player may take between 1 and 4 coins. (Or between 1 and N
coins.)
* The player who takes the last coin wins.
* There are 2 (or more) piles of coins. Each player may only take
coins from one pile.
I hope this helps. If you have any more questions or comments, write
back again.
- Doctor TWE, The Math Forum
http://mathforum.com/dr.math/

Date: 02/27/2002 at 16:07:59
From: Doctor Anthony
Subject: Re: NIM
The general approach to the game of NIM is described below.
The technique is to express the number of coins in each pile in binary
scale and add the coefficients of the various powers of 2. You then
remove as many coins from some pile or other as necessary to leave the
sum of the coefficients of each power of 2 an even number. When the
opponent draws he is bound to upset such an arrangement and you then
repeat the process. The only exception is that you must not leave an
even number of piles containing one coin each.
We can lay out the problem as follows, with three piles containing say
3, 4, and 5 coins respectively
Pile(1) Pile(2) Pile(3) 11 (pile(1) in binary)
------- ------- --------- 100 (pile(2) in binary)
*** **** ***** 101 (pile(3) in binary)
-----
212 = Sum of columns
We want the sum of each column to be even (or zero) and we can do this
by making pile(1) = 1 instead of 11. So we reduce pile(1) to 1. This
means we must take 2 coins from pile(1). We then get:
1
100
101
-----
202 This is now a winning position.
When the opponent takes any coins he is bound to leave some column
odd, and you can repeat the above calculation to get back to each
column having an even total. As mentioned before, you must not leave
an even number of piles with one coin in each (this does give an even
sum to the column 2^0), and in this case you must leave an ODD number
of piles with one coin in each pile.
- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/